To solve this, we use the concept of conditional probability. The probability of rolling a number greater than $7$ given that the number rolled is greater than $5$ can be written as:

$P(\text{roll > 7 | roll > 5}) = \frac{P(\text{roll > 7 and roll > 5})}{P(\text{roll > 5})}.$

Step 1: Determine the outcomes

• The die is 8-sided, so the possible outcomes are $1, 2, 3, 4, 5, 6, 7, 8$.
• Numbers greater than $5$ are $6, 7, 8$, so there are $3$ outcomes.
• Numbers greater than $7$ are $8$, so there is $1$ outcome.
• The intersection of "greater than $7$" and "greater than $5$" is simply numbers greater than $7$, i.e., $8$.

Step 2: Calculate probabilities

• $P(\text{roll > 5}) = \frac{\text{favorable outcomes (6, 7, 8)}}{\text{total outcomes (1, 2, ..., 8)}} = \frac{3}{8}$.
• $P(\text{roll > 7 and roll > 5}) = \frac{\text{favorable outcomes (8)}}{\text{total outcomes (1, 2, ..., 8)}} = \frac{1}{8}$.

Step 3: Calculate conditional probability

$P(\text{roll > 7 | roll > 5}) = \frac{P(\text{roll > 7 and roll > 5})}{P(\text{roll > 5})} = \frac{\frac{1}{8}}{\frac{3}{8}} = \frac{1}{3}.$

Final Answer:

The probability is:

$\boxed{\frac{1}{3}}$

Let me know if you'd like more details or have further questions! Here are five related questions to explore further:

1. What is the probability of rolling a number less than 4 on the same 8-sided die?
2. How does the probability change if the die is biased instead of fair?
3. Can you explain how conditional probability applies in other real-life scenarios?
4. What is the probability of rolling an even number given that the roll is greater than 3?
5. How would the solution change if the die had 10 sides instead of 8?

Tip: Conditional probability often involves narrowing down the sample space to focus only on relevant outcomes.